147347
Which one of the following is a pair of isotones?
1 ${ }^{37} \mathrm{Cl}_{17}$ and ${ }^{40} \mathrm{~K}_{18}$
2 ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$
3 ${ }^{40} \mathrm{~K}_{18}$ and ${ }^{40} \mathrm{Ar}_{19}$
4 ${ }^{13} \mathrm{~N}_{7}$ and ${ }^{14} \mathrm{~N}_{7}$
Explanation:
B Isotones are atoms or nuclei which have equal number of neutrons Mass number $(\mathrm{A})=\mathrm{Z}+\mathrm{N}$ Where, $Z=$ atomic number $=$ number of proton. $\mathrm{N}=$ no. of neutron No. of neutron $=$ mass number - atomic number No. of neutron in ${ }^{198} \mathrm{Hg}_{80}=198-80$ Similarly in ${ }^{197} \mathrm{Au}_{79}$ $=118$ Number of neutron $=$ mass number - atomic number $=197-79$ Number of neutrons $=118$ So, ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$ are isotones.
MHT-CET 2020
NUCLEAR PHYSICS
147348
The graph of $\ln \left(\frac{R}{R_{0}}\right)$ versus $\ln A$ is where $R$ is radius of a nucleus. $A$ is its mass number, and $R_{0}$ is constant
1 A straight line
2 A circle of radius $\mathrm{R}$
3 A parabola
4 An ellipse
Explanation:
A We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{A}^{1 / 3}$ Taking log both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ Let $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\mathrm{y}, \ln \mathrm{A}=\mathrm{x}$ Then $y=x / 3$. It is an equation of straight line passing through the origin.
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147349
A nucleus has mass number $A_{1}$ and volume $V_{1}$. Another nucleus has mass number $A_{2}$ and volume $V_{2}$. If the relation between the mass numbers is $A_{2}=3 A_{1}$, then $\frac{V_{1}}{V_{2}}=$
1 $(3)^{1 / 3}$
2 $\left(\frac{1}{3}\right)^{1 / 3}$
3 $\frac{1}{3}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C We know that, the radius of the nucleus given by, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\text { Volume }(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{~V}=\frac{4}{3} \pi\left(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\right)^{3}$ $\mathrm{~V}=\frac{4}{3} \pi \mathrm{AR}_{0}^{3}$ For volume $\mathrm{V}_{1}$, $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{A}_{1} \mathrm{R}_{0}^{3}$ For volume $\mathrm{V}_{2}$, $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{2} \mathrm{R}_{0}^{3}$ $\mathrm{~V}_{2}=\frac{4}{3} \pi\left(3 \mathrm{~A}_{1}\right) \mathrm{R}_{0}^{3} \quad\left(\therefore \mathrm{A}_{2}=3 \mathrm{~A}_{1}\right)$ $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{1} \times \mathrm{R}_{0}^{3} \times 3$ From equation (i) divided by (ii), we get - $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$
AP EAMCET-08.07.2022
NUCLEAR PHYSICS
147350
The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is
1 64
2 81
3 100
4 125
Explanation:
D Given that, $\mathrm{R}_{1}=4.8 \text { fermi } \mathrm{A}_{1}=64$ $\mathrm{R}_{2}=6 \text { fermi }$ We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=\left(\frac{4.8}{6}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=(0.8)^{3}$ $\frac{\mathrm{A}_{2}}{1}=\frac{64}{(0.8)^{3}}$ $\mathrm{~A}_{2}=125$
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147351
Read the following statements:
1 (A) and (D) only
2 (A) and (E) only
3 (B) and (E) only
4 (A) and (C) only
5 Density of the nucleus is independent of the mass number. Choose the correct option from the following options.
Explanation:
B Radius of nucleus, $R=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{\frac{1}{3}}$ Where $R_{o}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number of nucleus Thus, $\mathrm{R} \propto \mathrm{A}^{\frac{1}{3}}$ Volume of nucleus $(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{V} \propto \mathrm{R}^{3}$ or $\mathrm{V} \propto \mathrm{A}$ Now, the density of nucleus is given by $\rho=\frac{M}{V}=\frac{m A}{\frac{4}{3} \pi R_{0}{ }^{3} A}=\frac{m}{\frac{4}{3} \pi R^{3}}$ So, density is independent of 'A'
147347
Which one of the following is a pair of isotones?
1 ${ }^{37} \mathrm{Cl}_{17}$ and ${ }^{40} \mathrm{~K}_{18}$
2 ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$
3 ${ }^{40} \mathrm{~K}_{18}$ and ${ }^{40} \mathrm{Ar}_{19}$
4 ${ }^{13} \mathrm{~N}_{7}$ and ${ }^{14} \mathrm{~N}_{7}$
Explanation:
B Isotones are atoms or nuclei which have equal number of neutrons Mass number $(\mathrm{A})=\mathrm{Z}+\mathrm{N}$ Where, $Z=$ atomic number $=$ number of proton. $\mathrm{N}=$ no. of neutron No. of neutron $=$ mass number - atomic number No. of neutron in ${ }^{198} \mathrm{Hg}_{80}=198-80$ Similarly in ${ }^{197} \mathrm{Au}_{79}$ $=118$ Number of neutron $=$ mass number - atomic number $=197-79$ Number of neutrons $=118$ So, ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$ are isotones.
MHT-CET 2020
NUCLEAR PHYSICS
147348
The graph of $\ln \left(\frac{R}{R_{0}}\right)$ versus $\ln A$ is where $R$ is radius of a nucleus. $A$ is its mass number, and $R_{0}$ is constant
1 A straight line
2 A circle of radius $\mathrm{R}$
3 A parabola
4 An ellipse
Explanation:
A We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{A}^{1 / 3}$ Taking log both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ Let $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\mathrm{y}, \ln \mathrm{A}=\mathrm{x}$ Then $y=x / 3$. It is an equation of straight line passing through the origin.
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147349
A nucleus has mass number $A_{1}$ and volume $V_{1}$. Another nucleus has mass number $A_{2}$ and volume $V_{2}$. If the relation between the mass numbers is $A_{2}=3 A_{1}$, then $\frac{V_{1}}{V_{2}}=$
1 $(3)^{1 / 3}$
2 $\left(\frac{1}{3}\right)^{1 / 3}$
3 $\frac{1}{3}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C We know that, the radius of the nucleus given by, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\text { Volume }(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{~V}=\frac{4}{3} \pi\left(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\right)^{3}$ $\mathrm{~V}=\frac{4}{3} \pi \mathrm{AR}_{0}^{3}$ For volume $\mathrm{V}_{1}$, $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{A}_{1} \mathrm{R}_{0}^{3}$ For volume $\mathrm{V}_{2}$, $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{2} \mathrm{R}_{0}^{3}$ $\mathrm{~V}_{2}=\frac{4}{3} \pi\left(3 \mathrm{~A}_{1}\right) \mathrm{R}_{0}^{3} \quad\left(\therefore \mathrm{A}_{2}=3 \mathrm{~A}_{1}\right)$ $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{1} \times \mathrm{R}_{0}^{3} \times 3$ From equation (i) divided by (ii), we get - $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$
AP EAMCET-08.07.2022
NUCLEAR PHYSICS
147350
The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is
1 64
2 81
3 100
4 125
Explanation:
D Given that, $\mathrm{R}_{1}=4.8 \text { fermi } \mathrm{A}_{1}=64$ $\mathrm{R}_{2}=6 \text { fermi }$ We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=\left(\frac{4.8}{6}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=(0.8)^{3}$ $\frac{\mathrm{A}_{2}}{1}=\frac{64}{(0.8)^{3}}$ $\mathrm{~A}_{2}=125$
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147351
Read the following statements:
1 (A) and (D) only
2 (A) and (E) only
3 (B) and (E) only
4 (A) and (C) only
5 Density of the nucleus is independent of the mass number. Choose the correct option from the following options.
Explanation:
B Radius of nucleus, $R=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{\frac{1}{3}}$ Where $R_{o}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number of nucleus Thus, $\mathrm{R} \propto \mathrm{A}^{\frac{1}{3}}$ Volume of nucleus $(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{V} \propto \mathrm{R}^{3}$ or $\mathrm{V} \propto \mathrm{A}$ Now, the density of nucleus is given by $\rho=\frac{M}{V}=\frac{m A}{\frac{4}{3} \pi R_{0}{ }^{3} A}=\frac{m}{\frac{4}{3} \pi R^{3}}$ So, density is independent of 'A'
147347
Which one of the following is a pair of isotones?
1 ${ }^{37} \mathrm{Cl}_{17}$ and ${ }^{40} \mathrm{~K}_{18}$
2 ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$
3 ${ }^{40} \mathrm{~K}_{18}$ and ${ }^{40} \mathrm{Ar}_{19}$
4 ${ }^{13} \mathrm{~N}_{7}$ and ${ }^{14} \mathrm{~N}_{7}$
Explanation:
B Isotones are atoms or nuclei which have equal number of neutrons Mass number $(\mathrm{A})=\mathrm{Z}+\mathrm{N}$ Where, $Z=$ atomic number $=$ number of proton. $\mathrm{N}=$ no. of neutron No. of neutron $=$ mass number - atomic number No. of neutron in ${ }^{198} \mathrm{Hg}_{80}=198-80$ Similarly in ${ }^{197} \mathrm{Au}_{79}$ $=118$ Number of neutron $=$ mass number - atomic number $=197-79$ Number of neutrons $=118$ So, ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$ are isotones.
MHT-CET 2020
NUCLEAR PHYSICS
147348
The graph of $\ln \left(\frac{R}{R_{0}}\right)$ versus $\ln A$ is where $R$ is radius of a nucleus. $A$ is its mass number, and $R_{0}$ is constant
1 A straight line
2 A circle of radius $\mathrm{R}$
3 A parabola
4 An ellipse
Explanation:
A We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{A}^{1 / 3}$ Taking log both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ Let $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\mathrm{y}, \ln \mathrm{A}=\mathrm{x}$ Then $y=x / 3$. It is an equation of straight line passing through the origin.
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147349
A nucleus has mass number $A_{1}$ and volume $V_{1}$. Another nucleus has mass number $A_{2}$ and volume $V_{2}$. If the relation between the mass numbers is $A_{2}=3 A_{1}$, then $\frac{V_{1}}{V_{2}}=$
1 $(3)^{1 / 3}$
2 $\left(\frac{1}{3}\right)^{1 / 3}$
3 $\frac{1}{3}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C We know that, the radius of the nucleus given by, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\text { Volume }(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{~V}=\frac{4}{3} \pi\left(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\right)^{3}$ $\mathrm{~V}=\frac{4}{3} \pi \mathrm{AR}_{0}^{3}$ For volume $\mathrm{V}_{1}$, $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{A}_{1} \mathrm{R}_{0}^{3}$ For volume $\mathrm{V}_{2}$, $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{2} \mathrm{R}_{0}^{3}$ $\mathrm{~V}_{2}=\frac{4}{3} \pi\left(3 \mathrm{~A}_{1}\right) \mathrm{R}_{0}^{3} \quad\left(\therefore \mathrm{A}_{2}=3 \mathrm{~A}_{1}\right)$ $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{1} \times \mathrm{R}_{0}^{3} \times 3$ From equation (i) divided by (ii), we get - $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$
AP EAMCET-08.07.2022
NUCLEAR PHYSICS
147350
The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is
1 64
2 81
3 100
4 125
Explanation:
D Given that, $\mathrm{R}_{1}=4.8 \text { fermi } \mathrm{A}_{1}=64$ $\mathrm{R}_{2}=6 \text { fermi }$ We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=\left(\frac{4.8}{6}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=(0.8)^{3}$ $\frac{\mathrm{A}_{2}}{1}=\frac{64}{(0.8)^{3}}$ $\mathrm{~A}_{2}=125$
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147351
Read the following statements:
1 (A) and (D) only
2 (A) and (E) only
3 (B) and (E) only
4 (A) and (C) only
5 Density of the nucleus is independent of the mass number. Choose the correct option from the following options.
Explanation:
B Radius of nucleus, $R=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{\frac{1}{3}}$ Where $R_{o}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number of nucleus Thus, $\mathrm{R} \propto \mathrm{A}^{\frac{1}{3}}$ Volume of nucleus $(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{V} \propto \mathrm{R}^{3}$ or $\mathrm{V} \propto \mathrm{A}$ Now, the density of nucleus is given by $\rho=\frac{M}{V}=\frac{m A}{\frac{4}{3} \pi R_{0}{ }^{3} A}=\frac{m}{\frac{4}{3} \pi R^{3}}$ So, density is independent of 'A'
147347
Which one of the following is a pair of isotones?
1 ${ }^{37} \mathrm{Cl}_{17}$ and ${ }^{40} \mathrm{~K}_{18}$
2 ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$
3 ${ }^{40} \mathrm{~K}_{18}$ and ${ }^{40} \mathrm{Ar}_{19}$
4 ${ }^{13} \mathrm{~N}_{7}$ and ${ }^{14} \mathrm{~N}_{7}$
Explanation:
B Isotones are atoms or nuclei which have equal number of neutrons Mass number $(\mathrm{A})=\mathrm{Z}+\mathrm{N}$ Where, $Z=$ atomic number $=$ number of proton. $\mathrm{N}=$ no. of neutron No. of neutron $=$ mass number - atomic number No. of neutron in ${ }^{198} \mathrm{Hg}_{80}=198-80$ Similarly in ${ }^{197} \mathrm{Au}_{79}$ $=118$ Number of neutron $=$ mass number - atomic number $=197-79$ Number of neutrons $=118$ So, ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$ are isotones.
MHT-CET 2020
NUCLEAR PHYSICS
147348
The graph of $\ln \left(\frac{R}{R_{0}}\right)$ versus $\ln A$ is where $R$ is radius of a nucleus. $A$ is its mass number, and $R_{0}$ is constant
1 A straight line
2 A circle of radius $\mathrm{R}$
3 A parabola
4 An ellipse
Explanation:
A We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{A}^{1 / 3}$ Taking log both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ Let $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\mathrm{y}, \ln \mathrm{A}=\mathrm{x}$ Then $y=x / 3$. It is an equation of straight line passing through the origin.
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147349
A nucleus has mass number $A_{1}$ and volume $V_{1}$. Another nucleus has mass number $A_{2}$ and volume $V_{2}$. If the relation between the mass numbers is $A_{2}=3 A_{1}$, then $\frac{V_{1}}{V_{2}}=$
1 $(3)^{1 / 3}$
2 $\left(\frac{1}{3}\right)^{1 / 3}$
3 $\frac{1}{3}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C We know that, the radius of the nucleus given by, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\text { Volume }(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{~V}=\frac{4}{3} \pi\left(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\right)^{3}$ $\mathrm{~V}=\frac{4}{3} \pi \mathrm{AR}_{0}^{3}$ For volume $\mathrm{V}_{1}$, $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{A}_{1} \mathrm{R}_{0}^{3}$ For volume $\mathrm{V}_{2}$, $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{2} \mathrm{R}_{0}^{3}$ $\mathrm{~V}_{2}=\frac{4}{3} \pi\left(3 \mathrm{~A}_{1}\right) \mathrm{R}_{0}^{3} \quad\left(\therefore \mathrm{A}_{2}=3 \mathrm{~A}_{1}\right)$ $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{1} \times \mathrm{R}_{0}^{3} \times 3$ From equation (i) divided by (ii), we get - $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$
AP EAMCET-08.07.2022
NUCLEAR PHYSICS
147350
The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is
1 64
2 81
3 100
4 125
Explanation:
D Given that, $\mathrm{R}_{1}=4.8 \text { fermi } \mathrm{A}_{1}=64$ $\mathrm{R}_{2}=6 \text { fermi }$ We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=\left(\frac{4.8}{6}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=(0.8)^{3}$ $\frac{\mathrm{A}_{2}}{1}=\frac{64}{(0.8)^{3}}$ $\mathrm{~A}_{2}=125$
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147351
Read the following statements:
1 (A) and (D) only
2 (A) and (E) only
3 (B) and (E) only
4 (A) and (C) only
5 Density of the nucleus is independent of the mass number. Choose the correct option from the following options.
Explanation:
B Radius of nucleus, $R=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{\frac{1}{3}}$ Where $R_{o}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number of nucleus Thus, $\mathrm{R} \propto \mathrm{A}^{\frac{1}{3}}$ Volume of nucleus $(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{V} \propto \mathrm{R}^{3}$ or $\mathrm{V} \propto \mathrm{A}$ Now, the density of nucleus is given by $\rho=\frac{M}{V}=\frac{m A}{\frac{4}{3} \pi R_{0}{ }^{3} A}=\frac{m}{\frac{4}{3} \pi R^{3}}$ So, density is independent of 'A'
147347
Which one of the following is a pair of isotones?
1 ${ }^{37} \mathrm{Cl}_{17}$ and ${ }^{40} \mathrm{~K}_{18}$
2 ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$
3 ${ }^{40} \mathrm{~K}_{18}$ and ${ }^{40} \mathrm{Ar}_{19}$
4 ${ }^{13} \mathrm{~N}_{7}$ and ${ }^{14} \mathrm{~N}_{7}$
Explanation:
B Isotones are atoms or nuclei which have equal number of neutrons Mass number $(\mathrm{A})=\mathrm{Z}+\mathrm{N}$ Where, $Z=$ atomic number $=$ number of proton. $\mathrm{N}=$ no. of neutron No. of neutron $=$ mass number - atomic number No. of neutron in ${ }^{198} \mathrm{Hg}_{80}=198-80$ Similarly in ${ }^{197} \mathrm{Au}_{79}$ $=118$ Number of neutron $=$ mass number - atomic number $=197-79$ Number of neutrons $=118$ So, ${ }^{198} \mathrm{Hg}_{80}$ and ${ }^{197} \mathrm{Au}_{79}$ are isotones.
MHT-CET 2020
NUCLEAR PHYSICS
147348
The graph of $\ln \left(\frac{R}{R_{0}}\right)$ versus $\ln A$ is where $R$ is radius of a nucleus. $A$ is its mass number, and $R_{0}$ is constant
1 A straight line
2 A circle of radius $\mathrm{R}$
3 A parabola
4 An ellipse
Explanation:
A We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ $\frac{\mathrm{R}}{\mathrm{R}_{0}}=\mathrm{A}^{1 / 3}$ Taking log both side $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\frac{1}{3} \ln \mathrm{A}$ Let $\ln \left(\frac{\mathrm{R}}{\mathrm{R}_{0}}\right)=\mathrm{y}, \ln \mathrm{A}=\mathrm{x}$ Then $y=x / 3$. It is an equation of straight line passing through the origin.
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147349
A nucleus has mass number $A_{1}$ and volume $V_{1}$. Another nucleus has mass number $A_{2}$ and volume $V_{2}$. If the relation between the mass numbers is $A_{2}=3 A_{1}$, then $\frac{V_{1}}{V_{2}}=$
1 $(3)^{1 / 3}$
2 $\left(\frac{1}{3}\right)^{1 / 3}$
3 $\frac{1}{3}$
4 $\frac{1}{\sqrt{3}}$
Explanation:
C We know that, the radius of the nucleus given by, $\mathrm{R}=\mathrm{R}_{0} \mathrm{~A}^{1 / 3}$ $\text { Volume }(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{~V}=\frac{4}{3} \pi\left(\mathrm{R}_{0} \mathrm{~A}^{1 / 3}\right)^{3}$ $\mathrm{~V}=\frac{4}{3} \pi \mathrm{AR}_{0}^{3}$ For volume $\mathrm{V}_{1}$, $\mathrm{V}_{1}=\frac{4}{3} \pi \mathrm{A}_{1} \mathrm{R}_{0}^{3}$ For volume $\mathrm{V}_{2}$, $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{2} \mathrm{R}_{0}^{3}$ $\mathrm{~V}_{2}=\frac{4}{3} \pi\left(3 \mathrm{~A}_{1}\right) \mathrm{R}_{0}^{3} \quad\left(\therefore \mathrm{A}_{2}=3 \mathrm{~A}_{1}\right)$ $\mathrm{V}_{2}=\frac{4}{3} \pi \mathrm{A}_{1} \times \mathrm{R}_{0}^{3} \times 3$ From equation (i) divided by (ii), we get - $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{1}{3}$
AP EAMCET-08.07.2022
NUCLEAR PHYSICS
147350
The radius of an atomic nucleus of mass number 64 is 4.8 fermi. Then the mass number of another atomic nucleus of radius 6 fermi is
1 64
2 81
3 100
4 125
Explanation:
D Given that, $\mathrm{R}_{1}=4.8 \text { fermi } \mathrm{A}_{1}=64$ $\mathrm{R}_{2}=6 \text { fermi }$ We know that, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{\frac{1}{3}}$ $\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=\left(\frac{4.8}{6}\right)^{3}$ $\frac{64}{\mathrm{~A}_{2}}=(0.8)^{3}$ $\frac{\mathrm{A}_{2}}{1}=\frac{64}{(0.8)^{3}}$ $\mathrm{~A}_{2}=125$
AP EAMCET-04.07.2022
NUCLEAR PHYSICS
147351
Read the following statements:
1 (A) and (D) only
2 (A) and (E) only
3 (B) and (E) only
4 (A) and (C) only
5 Density of the nucleus is independent of the mass number. Choose the correct option from the following options.
Explanation:
B Radius of nucleus, $R=\mathrm{R}_{\mathrm{o}} \mathrm{A}^{\frac{1}{3}}$ Where $R_{o}=1.2 \times 10^{-15} \mathrm{~m}$ $\mathrm{A}=$ mass number of nucleus Thus, $\mathrm{R} \propto \mathrm{A}^{\frac{1}{3}}$ Volume of nucleus $(\mathrm{V})=\frac{4}{3} \pi \mathrm{R}^{3}$ $\mathrm{V} \propto \mathrm{R}^{3}$ or $\mathrm{V} \propto \mathrm{A}$ Now, the density of nucleus is given by $\rho=\frac{M}{V}=\frac{m A}{\frac{4}{3} \pi R_{0}{ }^{3} A}=\frac{m}{\frac{4}{3} \pi R^{3}}$ So, density is independent of 'A'
